Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $r = \dfrac{2n - 2}{n + 10} \times \dfrac{n^2 + 14n + 40}{n^2 - n} $
Explanation: First factor the quadratic. $r = \dfrac{2n - 2}{n + 10} \times \dfrac{(n + 10)(n + 4)}{n^2 - n} $ Then factor out any other terms. $r = \dfrac{2(n - 1)}{n + 10} \times \dfrac{(n + 10)(n + 4)}{n(n - 1)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ 2(n - 1) \times (n + 10)(n + 4) } { (n + 10) \times n(n - 1) } $ $r = \dfrac{ 2(n - 1)(n + 10)(n + 4)}{ n(n + 10)(n - 1)} $ Notice that $(n - 1)$ and $(n + 10)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ 2\cancel{(n - 1)}(n + 10)(n + 4)}{ n\cancel{(n + 10)}(n - 1)} $ We are dividing by $n + 10$ , so $n + 10 \neq 0$ Therefore, $n \neq -10$ $r = \dfrac{ 2\cancel{(n - 1)}\cancel{(n + 10)}(n + 4)}{ n\cancel{(n + 10)}\cancel{(n - 1)}} $ We are dividing by $n - 1$ , so $n - 1 \neq 0$ Therefore, $n \neq 1$ $r = \dfrac{2(n + 4)}{n} ; \space n \neq -10 ; \space n \neq 1 $